Lemma 82.2.4. In Situation 82.2.1 assume $B$ is Jacobson and that $\delta (b) = 0$ for every closed point $b$ of $|B|$. Let $X/B$ be good. If $Z \subset X$ is an integral closed subspace with generic point $\xi \in |Z|$, then the following integers are the same:
$\delta (\xi ) = \delta _{X/B}(\xi )$,
$\dim (|Z|)$,
$\text{codim}(\{ z\} , |Z|)$ for $z \in |Z|$ closed,
the dimension of the local ring of $Z$ at $z$ for $z \in |Z|$ closed, and
$\dim (\mathcal{O}_{Z, \overline{z}})$ for $z \in |Z|$ closed.
Proof.
Let $X$, $Z$, $\xi $ be as in the lemma. Since $X$ is locally of finite type over $B$ we see that $X$ is Jacobson, see Decent Spaces, Lemma 68.23.1. Hence $X_{\text{ft-pts}} \subset |X|$ is the set of closed points by Decent Spaces, Lemma 68.23.3. Given a chain $T_0 \supset \ldots \supset T_ e$ of irreducible closed subsets of $|Z|$ we have $T_ e \cap X_{\text{ft-pts}}$ nonempty by Morphisms of Spaces, Lemma 67.25.6. Thus we can always assume such a chain ends with $T_ e = \{ z\} $ for some $z \in |Z|$ closed. It follows that $\dim (Z) = \sup _ z \text{codim}(\{ z\} , |Z|)$ where $z$ runs over the closed points of $|Z|$. We have $\text{codim}(\{ z\} , Z) = \delta (\xi ) - \delta (z)$ by Topology, Lemma 5.20.2. By Morphisms of Spaces, Lemma 67.25.4 the image of $z$ is a finite type point of $B$, i.e., a closed point of $|B|$. By Morphisms of Spaces, Lemma 67.33.4 the transcendence degree of $z/b$ is $0$. We conclude that $\delta (z) = \delta (b) = 0$ by assumption. Thus we obtain equality
\[ \dim (|Z|) = \text{codim}(\{ z\} , Z) = \delta (\xi ) \]
for all $z \in |Z|$ closed. Finally, we have that $\text{codim}(\{ z\} , Z)$ is equal to the dimension of the local ring of $Z$ at $z$ by Decent Spaces, Lemma 68.20.2 which in turn is equal to $\dim (\mathcal{O}_{Z, \overline{z}})$ by Properties of Spaces, Lemma 66.22.4.
$\square$
Comments (0)